Question

Figure shows three particles with charges $q_1=2Q,q_2=-2Q$ and $q_3=-4Q$, each a distance $d$ from the origin. Find the net electric field $\vec{E}$ at the origin.

• A. $\frac{2.56Q}{4\pi {\epsilon }_0{d}^2}$ towards + ve x-axis
• B. $\frac{6.932Q}{4\pi {\epsilon }_0{d}^2}$ towards + ve x-axis
• C. $\frac{6.93Q}{4\pi {\epsilon }_0{d}^2}$ towards - ve x-axis
• D. Zero

Answer 1

Answer: (B)

The given, $q_1=2Q,q_2=-2Q,q_3=-4Q$
The net electric field $E$ at the origin
$E=\left(\frac{1}{4\pi {\epsilon }_0}\cdot \frac{q_1}{d^2}+\frac{1}{4\pi {\epsilon }_0}\cdot \frac{q_2}{d^2}+\frac{1}{4\pi {\epsilon }_0}\cdot \frac{q_3}{d^2}\right)\cos \theta$
$=\left(\frac{1}{4\pi {\epsilon }_0}\cdot \frac{2Q}{d^2}+\frac{1}{4\pi {\epsilon }_0}\cdot \frac{2Q}{d^2}+\frac{1}{4\pi {\epsilon }_0}\cdot \frac{4Q}{d^2}\right)\cos \theta$
$=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{2\times 4Q}{d^2}\times \frac{\sqrt{3}}{2}$
$=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{8\times 1.732Q}{d^2\times 2}=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{6.928Q}{d^2}$
$=\frac{6.932Q}{4\pi {\epsilon }_0\cdot d^2}$ towards $+$ ve $x$-axis.